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Java memory Mapped File from Windows system

I'm trying something new,
There is an application who send data to a memory-mapped file located at Local\MemFileName
I would like to read it in java,
I tried some tutorials like https://www.baeldung.com/java-mapped-byte-buffer, https://howtodoinjava.com/java7/nio/memory-mapped-files-mappedbytebuffer/
But all seem to read a file in JVM, or I did not understand...
How can I read the content of the file Located in windows system Local\MemFileName
Thanks!
Following: Example code of what i tried
public class Main {
private static final String IRSDKMEM_MAP_FILE_NAME = StringEscapeUtils.unescapeJava("Local\\IRSDKMemMapFileName");
private static final String IRSDKDATA_VALID_EVENT = StringEscapeUtils.unescapeJava("Local\\IRSDKDataValidEvent");
public static final CharSequence charSequence = "Local\\IRSDKMemMapFileName";
public static void main(String[] args) throws IOException, InterruptedException {
System.out.println(charSequence);
try (RandomAccessFile file = new RandomAccessFile(new File(IRSDKMEM_MAP_FILE_NAME), "r")) {
//Get file channel in read-only mode
FileChannel fileChannel = file.getChannel();
//Get direct byte buffer access using channel.map() operation
MappedByteBuffer buffer = fileChannel.map(FileChannel.MapMode.READ_ONLY, 0, fileChannel.size());
// the buffer now reads the file as if it were loaded in memory.
System.out.println("Loaded " + buffer.isLoaded()); //prints false
System.out.println("capacity" + buffer.capacity()); //Get the size based on content size of file
//You can read the file from this buffer the way you like.
for (int i = 0; i < buffer.limit(); i++) {
System.out.println((char) buffer.get()); //Print the content of file
}
}
}
}

To read a memory mapped file:
Open a FileChannel on the file using FileChannel.open.
Invoke the map method on the FileChannel to create a MappedByteBuffer covering the area of the file you want to read.
Read the data from the MappedByteBuffer.

The solution for me was to use a WindowsService.class implementing methods from JNA library, as you can see:
My Library
With this I could open a file mapped in Windows system.
All the previous answers was correct for a file accessible from JVM, but from outside the JVM it was impossible.
Thanks !

Load text file to memory in Java

I have wiki.txt file and its size is 50 MB.
I need to do several things on the file and so I thought that the best way in terms of performance is to load the file to memory, is that correct?
This is the code that I written:
File file = new File("wiki.txt");
FileInputStream fileInputStream = new FileInputStream(file);
FileChannel fileChannel = fileInputStream.getChannel();
MappedByteBuffer mapByteBuffer = fileChannel.map(FileChannel.MapMode.READ_ONLY, 0, file.length());
System.out.println((char)mapByteBuffer.get());
I get error on this code: mapByteBuffer.get().
I tried the get() function a few options but all of them I get error and didn't even get an error on e.getMessage() I just got null.
Another important thing to note, my text file contains English words and actions I need to do is search, if expressed is exist in this text file.
Thank you.

I would suggest using a MemoryMappedFile, to read the file directly from the disk instead of loading it in memory.
RandomAccessFile file = new RandomAccessFile("wiki.txt", "r");
FileChannel channel = file.getChannel();
MappedByteBuffer buf = channel.map(FileChannel.MapMode.READ_WRITE, 0, 1024*50);
And then you can read the buffer as usual.

My answers for point (1):
It depends on what you want to do with the file. If your processing doesn't involve rewind operation (looking what was read behind/before), it's best to just read as a stream and process it in one go (instead of loading all into memory).
Even if you need random access across the file, you may also be interested in doing block file operation, because your solution may not scale well when the file size change to bigger size.
RandomAccessFile if you are on Java 1.4 or above.
For random access, the operating system usually handles the file buffer caching quite well you don't have to handle yourself.

It is important to read the whole error, not just the message. Often the real information is in the exception's name not the text associated with it.
You will get an error if the file is empty as there is no first byte.
Note: the approach you are using assumes ASCII 7-bit characters. If you want to assume ISO-8859-1 characters you can use (char) (byteBuffer.get() & 0xFF)
However, if you have plan text you may find that using strings is simpler to use and not much slower. e.g. you can read a 50 MB file as text in less than a second. I would only use a memory mapped file if this is far too long.

I would suggest to use BufferedReader. It is much faster and requires relatively less resources.
First read number of lines:
InputStream is = new BufferedInputStream(new FileInputStream(filename));
byte[] chars = new byte[1024];
int numberOfChars = 0;
while ((numberOfChars = is.read(chars)) != -1)
{
for (int i = 0; i < numberOfChars; ++i)
{
if (chars[i] == '\n' && numberOfChars - i != 1)
{
++count;
}
}
}
count++
return count; // number of lines
Then read the lines:
BufferedReader in = new BufferedReader(new FileReader(fileName));
for (int i = 0; i < endLine; i++)
{
String oneLine = in.readLine();
}
In this strings you can even do search for what you need.

Write to file at particular index

I need to write to a file at a particular index position. BufferedWriter and PrintWriter does not allow writing to an index. How do I go about achieving this?
Basically what i want to do is if a file contains an empty line at the EOF then i need to write at that position, else insert a new line and write. Copying the contents of the file to a temporary file and then deleting the original file and then again renaming the temporary file to the original file's name isn't an option.
Thanks

You need to use a RandomAccessFile.
Using this class, you can go to a specific location using the seek(long) method and write using the different write methods.
In the case of your particuliar problem, the best solution weems to be
to use a RandomAccessFile and navigate to the end of your file. Check if this is a new line, write, close.

Given is the method to write content at particular position.
Lets say my file is Test.txt and content is as follow
Hello
How are you
Today is Monday
now you want to write "hi" after hello. So the offset for "hi" will be "5".
Method is :
filename = "test.txt";
offset = 5;
byte[] content = ("\t hi").getBytes();
private void insert(String filename, long offset, byte[] content) throws IOException {
RandomAccessFile r = new RandomAccessFile(filename, "rw");
RandomAccessFile rtemp = new RandomAccessFile(filename+"Temp", "rw");
long fileSize = r.length();
FileChannel sourceChannel = r.getChannel();
FileChannel targetChannel = rtemp.getChannel();
sourceChannel.transferTo(offset, (fileSize - offset), targetChannel);
sourceChannel.truncate(offset);
r.seek(offset);
r.write(content);
long newOffset = r.getFilePointer();
targetChannel.position(0L);
sourceChannel.transferFrom(targetChannel, newOffset, (fileSize - offset));
sourceChannel.close();
targetChannel.close();
rtemp.close();
r.close();
}
The output will be:
Hello hi
How are you
Today is Monday

Writing in the beginning of a text file Java

I need to write something into a text file's beginning. I have a text file with content and i want write something before this content. Say i have;
Good afternoon sir,how are you today?
I'm fine,how are you?
Thanks for asking,I'm great
After modifying,I want it to be like this:
Page 1-Scene 59
25.05.2011
Good afternoon sir,how are you today?
I'm fine,how are you?
Thanks for asking,I'm great
Just made up the content :) How can i modify a text file like this way?

You can't really modify it that way - file systems don't generally let you insert data in arbitrary locations - but you can:
Create a new file
Write the prefix to it
Copy the data from the old file to the new file
Move the old file to a backup location
Move the new file to the old file's location
Optionally delete the old backup file

Just in case it will be useful for someone here is full source code of method to prepend lines to a file using Apache Commons IO library. The code does not read whole file into memory, so will work on files of any size.
public static void prependPrefix(File input, String prefix) throws IOException {
LineIterator li = FileUtils.lineIterator(input);
File tempFile = File.createTempFile("prependPrefix", ".tmp");
BufferedWriter w = new BufferedWriter(new FileWriter(tempFile));
try {
w.write(prefix);
while (li.hasNext()) {
w.write(li.next());
w.write("\n");
}
} finally {
IOUtils.closeQuietly(w);
LineIterator.closeQuietly(li);
}
FileUtils.deleteQuietly(input);
FileUtils.moveFile(tempFile, input);
}

I think what you want is random access. Check out the related java tutorial. However, I don't believe you can just insert data at an arbitrary point in the file; If I recall correctly, you'd only overwrite the data. If you wanted to insert, you'd have to have your code
copy a block,
overwrite with your new stuff,
copy the next block,
overwrite with the previously copied block,
return to 3 until no more blocks

As #atk suggested, java.nio.channels.SeekableByteChannel is a good interface. But it is available from 1.7 only.
Update : If you have no issue using FileUtils then use
String fileString = FileUtils.readFileToString(file);

This isn't a direct answer to the question, but often files are accessed via InputStreams. If this is your use case, then you can chain input streams via SequenceInputStream to achieve the same result. E.g.
InputStream inputStream = new SequenceInputStream(new ByteArrayInputStream("my line\n".getBytes()), new FileInputStream(new File("myfile.txt")));

I will leave it here just in case anyone need
ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
try (FileInputStream fileInputStream1 = new FileInputStream(fileName1);
FileInputStream fileInputStream2 = new FileInputStream(fileName2)) {
while (fileInputStream2.available() > 0) {
byteArrayOutputStream.write(fileInputStream2.read());
}
while (fileInputStream1.available() > 0) {
byteArrayOutputStream.write(fileInputStream1.read());
}
}
try (FileOutputStream fileOutputStream = new FileOutputStream(fileName1)) {
byteArrayOutputStream.writeTo(fileOutputStream);
}

Inserting text into an existing file via Java

I would like to create a simple program (in Java) which edits text files - particularly one which performs inserting arbitrary pieces of text at random positions in a text file. This feature is part of a larger program I am currently writing.
Reading the description about java.util.RandomAccessFile, it appears that any write operations performed in the middle of a file would actually overwrite the exiting content. This is a side-effect which I would like to avoid (if possible).
Is there a simple way to achieve this?
Thanks in advance.

Okay, this question is pretty old, but FileChannels exist since Java 1.4 and I don't know why they aren't mentioned anywhere when dealing with the problem of replacing or inserting content in files. FileChannels are fast, use them.
Here's an example (ignoring exceptions and some other stuff):
public void insert(String filename, long offset, byte[] content) {
RandomAccessFile r = new RandomAccessFile(new File(filename), "rw");
RandomAccessFile rtemp = new RandomAccessFile(new File(filename + "~"), "rw");
long fileSize = r.length();
FileChannel sourceChannel = r.getChannel();
FileChannel targetChannel = rtemp.getChannel();
sourceChannel.transferTo(offset, (fileSize - offset), targetChannel);
sourceChannel.truncate(offset);
r.seek(offset);
r.write(content);
long newOffset = r.getFilePointer();
targetChannel.position(0L);
sourceChannel.transferFrom(targetChannel, newOffset, (fileSize - offset));
sourceChannel.close();
targetChannel.close();
}

Well, no, I don't believe there is a way to avoid overwriting existing content with a single, standard Java IO API call.
If the files are not too large, just read the entire file into an ArrayList (an entry per line) and either rewrite entries or insert new entries for new lines.
Then overwrite the existing file with new content, or move the existing file to a backup and write a new file.
Depending on how sophisticated the edits need to be, your data structure may need to change.
Another method would be to read characters from the existing file while writing to the edited file and edit the stream as it is read.

If Java has a way to memory map files, then what you can do is extend the file to its new length, map the file, memmove all the bytes down to the end to make a hole and write the new data into the hole.
This works in C. Never tried it in Java.
Another way I just thought of to do the same but with random file access.
Seek to the end - 1 MB
Read 1 MB
Write that to original position + gap size.
Repeat for each previous 1 MB working toward the beginning of the file.
Stop when you reach the desired gap position.
Use a larger buffer size for faster performance.

You can use following code:
BufferedReader reader = null;
BufferedWriter writer = null;
ArrayList list = new ArrayList();
try {
reader = new BufferedReader(new FileReader(fileName));
String tmp;
while ((tmp = reader.readLine()) != null)
list.add(tmp);
OUtil.closeReader(reader);
list.add(0, "Start Text");
list.add("End Text");
writer = new BufferedWriter(new FileWriter(fileName));
for (int i = 0; i < list.size(); i++)
writer.write(list.get(i) + "\r\n");
} catch (Exception e) {
e.printStackTrace();
} finally {
OUtil.closeReader(reader);
OUtil.closeWriter(writer);
}

I don't know if there's a handy way to do it straight otherwise than
read the beginning of the file and write it to target
write your new text to target
read the rest of the file and write it to target.
About the target : You can construct the new contents of the file in memory and then overwrite the old content of the file if the files handled aren't so big. Or you can write the result to a temporary file.
The thing would probably be easiest to do with streams, RandomAccessFile doesn't seem to be meant for inserting in the middle (afaik). Check the tutorial if you need.

I believe the only way to insert text into an existing text file is to read the original file and write the content in a temporary file with the new text inserted. Then erase the original file and rename the temporary file to the original name.
This example is focused on inserted a single line into an existing file, but still maybe of use to you.

If it is a text file,,,,Read the existing file in StringBuffer and append the new content in the same StringBuffer now u can write the SrtingBuffer on file. so now the file contains both the existing and new text.

As #xor_eq answer's edit queue is full, here in a new answer a more documented and slightly improved version of his:
public static void insert(String filename, long offset, byte[] content) throws IOException {
File temp = Files.createTempFile("insertTempFile", ".temp").toFile(); // Create a temporary file to save content to
try (RandomAccessFile r = new RandomAccessFile(new File(filename), "rw"); // Open file for read & write
RandomAccessFile rtemp = new RandomAccessFile(temp, "rw"); // Open temporary file for read & write
FileChannel sourceChannel = r.getChannel(); // Channel of file
FileChannel targetChannel = rtemp.getChannel()) { // Channel of temporary file
long fileSize = r.length();
sourceChannel.transferTo(offset, (fileSize - offset), targetChannel); // Copy content after insert index to
// temporary file
sourceChannel.truncate(offset); // Remove content past insert index from file
r.seek(offset); // Goto back of file (now insert index)
r.write(content); // Write new content
long newOffset = r.getFilePointer(); // The current offset
targetChannel.position(0L); // Goto start of temporary file
sourceChannel.transferFrom(targetChannel, newOffset, (fileSize - offset)); // Copy all content of temporary
// to end of file
}
Files.delete(temp.toPath()); // Delete the temporary file as not needed anymore
}